Final powerpoint decision science

TRANSPORTATION(SPECIALLY STRUCTURED LINEAR PROGRAMS)

SUBMITTED BY:

Transportation 1

Escober, Markde la Cruz, Edghan Bryan

Transportation 2

Kali, Norhasim Lagasca, MichelleRonidel, Roderick

TRANSPORTATION AND ASSIGNMENT PROBLEM

Transportation problem is concerned with selecting routes in a product – distribution networks among manufacturing plants to distribution warehouse or among regional warehouses to local distribution outlets.

Assignment problem involves assigning employees to tasks, salesperson to towns, contract to bidders or jobs to plants.

TRANSPORTATION METHOD

The Transportation Problem:

(demand equals supply)

Let us consider the case of the Bulacan Gravel Company , which has received a contract to supply gravel for three road projects located in the towns of Marilao, Bocaue and Balagtas. Construction engineers have estimated the amount of gravel which be needed at three construction projects:

Weekly Requirement

Project Location Truckloads

A Marilao 72

B Bocaue 102

C Balagtas 41

Total 215

The Transportation Problem:

(demand equals supply)

The Bulacan Gravel Company has three gravel plants located in the towns of San Rafael, San Ildefonso and San Miguel. The gravel required for the construction projects can be supplied by these three plants. Bulacan’s chief dispatcher has calculated the amounts of gravel which can be supplied by each plant:

The company has computed the delivery costs from each plant to each project site.

Cost per Truckload

From To project A To project B To project C

Plant W P 4 P 8 P 8

Plant X 16 24 16

Plant Y 8 16 24

Amount Available/Week

Plant Location Truckloads

W San Rafael 56

X San Ildefonso 82

Y San Miguel 77

Total 215

Gravel plants, road construction projects and transportation costs for Bulacan Gravel

Company

Project A

(Marilao)

72 loads required

Project B

(Bocaue)

102 loads

required

Project C

(Balagtas)

41 loads

required

Plant X ( San Ildefonso)

82 loads available

Plant W (San Rafael)

56 loads available

Plant Y (San Miguel)

77 loads available

P4

P24

P16

P16P8

P24

P16

P8

P8

Gravel plants, road construction projects and transportation costs for Bulacan Gravel

Company

The objective is to minimize the total transportation cost:

There are three “origin constraints” which say that Bulacan cannot ship out more gravel than they have:

There are three “destination constraints”, which say that each project must receive the gravel it requires:

4WA + 8WB + 8WC + 16XA + 24XB + 16XC + 8YA + 16YB + 24YC

WA + WB + WC < 56: Plant W

XA + XB + XC < 82 : Plant X

YA + YB + YC < 77: Plant Y

WA + XA + YA > 72: Project A

WB + XB + YB > 102: Project B

WC +XC + YC > 41:Project C

Using it all together, we get the final linear programming model:

minimize

4WA + 8WB + 8WC + 16XA + 24XB + 16XC + 8YA + 16YB + 24YC

subject to

WA+ WB+ WC < 56: Plant W

XA+ XB+ XC < 82: Plant X

YA+ YB+ YC< 77: Plant Y

WA + XA + YA > 72: Project A

WB + XB + YB>102: Project B

WC + XC + YC> 41: Project C

All Variables > 0

STEP 1: Set up the transportation table

STEP 2: Develop an initial solution

STEP 3: Test the solution for improvement

STEP 4: Develop the improved solution

Project A

Plant Y

Plant X

Project B

Project C

Plant W

Plant Capacit

y

ProjectRequireme

nts

56

72 102 41

77

82

FromTo

215215

A B

C

D

E

From

ToProject A Project B Project C

Plant Capacity

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102 41

56

82

77

215215

4 8

16

824

16

8

24

16

WA WB WC

XA

YA

XB XC

YB YC

X1X2

X3

X4 X5 X6

X7 X8 X9

+-

4

X1

WA12

3

From


Plant Capacity

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102 41

56

82

77

215215

4 8

16

824

16

8

24

16

WA WB WC

XA

YA

XB XC

YB YC

56

16 66

36 41

From To Quantity,

Plant Project truckloads/week

W A 56

X A 16

X B 66

Y B 36

Y C 41

215

Used Squares = Total Rim Requirements – 1

5 = 6 - 1

Total Cost of the Initial Solution

Is this the best solution?

Source - destination Quantity Unit Total

Combination Shipped x Cost = Cost

WA 56 P 4 224.00

XA 16 16 256.00

XB 66 24 1,584.00

YB 36 16 576.00

YC 41 24 984.00

Total Transporation Cost P 3,624.00

Choose the unused square to be evaluated

Beginning with the selected unused square, traced a closed path (moving horizontally and vertically only)

Assign plus (+) and minus (-) signs alternately at each corner square of the closed path

Determine the net change in costs as a result of the changes made in tracing the path

Repeat the above steps until an improvement index has been determined for each unused squares

From


Plant Capacity

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102 41

56

82

77

215215

4 8

16

824

16

8

24

16

WA WB WC

XA

YA

XB XC

YB YC

56

16 66

36 41

_

+ _

+

Addition to cost: From plant W to project B P 8

From plant X to project A 16 P24

Reduction to cost: From plant W to project A P 4

From plant X to project B 24 28

- P 4

Improvement index for square WB = WB – WA + XA – XB

= P8 - P4 + P16 - P24

WB = -P4

Improvement index for WC = WC – WA + XA – XB + YB – YC

= P8 – 4P + P16 -P24 + P16 - P 24

WC = - P 12

Improvement index for XC= XC – XB + YB – YC

= P16 - P24 + P16 - P24

XC = -P16

Improvement index for YA = YA - XA + XB – YB

= P8 - P16 + P24 - P16

YA = P 0

From


Plant Capacity

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102 41

56

82

77

215215

4 8

16

824

16

8

24

16

WA WB WC

XA

YA

XB XC

YB YC

56

16 66

36 41

-4 -12

-16

0

`

24 16

2416

66

36

41

_

+ _

+

XB XC

YB YC

`

41

XB XC

YB YC

25

77 0

0 + 4166 - 41

36 + 41 41 - 41

From


Plant Capacity

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102

n

41

56

82

77

215215

4 8

16

824

16

8

24

16

WA WB WC

XA

YA

XB XC

YB YC

56

16 25

77

41

Shipping Quantity Unit Total

Assignments Shipped x Cost = Cost

WA 56 P 4 P224

XA 16 16 256

XB 25 24 600

XC 41 16 656

YB 77 16 1,232

P 2,968

Unused Closed Computation of

Squares path Improvement Index

WB + WB - WA + XA - XB + 8 - 4 + 16 - 24 = - 4

WC + WC - WA + XA - XC + 8 - 4 + 16 - 16 = + 4

YA + YA - XA + XB - YB + 8 - 16 + 24 - 16 = 0

YC + YC - XC + XB - YB + 24 - 16 + 24 -16 = + 16

`

25

WA WB

XA XB

31

41 0

0 + 2556 - 25

16 + 25 = 25 - 25

From


Plant Capacity

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102

n

41

56

82

77

215215

4 8

16

824

16

8

24

16

WA WB WC

XA

YA

XB XC

YB YC

31

41

25

77

41



WA 31 P 4 P124

WB 25 8 200

XA 41 16 656

XC 41 16 656

YB 77 16 1,232

P 2,868



WC +WC - WA + XA - XC +8 – 4 + 16 – 16 = + 4

XB +XB - WB + WA - XA +24 – 8 + 4 – 16 = + 4

YA +YA - WA + WB – YB +8 – 4 + 8 – 16 = - 4

YC +YC - YB + WB - WA + XA - XC + 24 – 16 + 8 – 4 + 16 – 16 = + 12

From


Plant Capacity

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102

n

41

56

82

77

215215

4 8

16

824

16

8

24

16

WA WB WC

XA

YA

XB XC

YB YC

31

41

56

46

41



WA +WA – YA + YB - WB +4 – 8 + 16 – 8 = + 4

WC +WC – XC + XA – YA + YB - WB +8 – 16 + 16 – 8 + 16 – 8 = + 8

XB +XB – YB + YA - XA +24 – 16 + 8 – 16 = 0

YC +YC – XC + XA - YA +24 – 16 + 16 – 8 = +16



WB 56 P 8 P448

XA 41 16 656

XC 41 16 656

YA 31 8 248

YB 46 16 736

P 2,744Total Transportation Cost

R i = value assigned to row i

K j = value assigned to row j

C i j = cost in square ij ( the square at the intersection

of row i and column j

Project A

Project B

Project C

Plant Capacit

y

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102 41

56

82

77

R2

R3

Ri

Kj K1 K2 K3

R1

215

215

56

16

66

36

41

4 8 8

16

24

16

16

24

8

WA WB WC

XA XB XC

YA YB YC

From

To

R1R2

R2

R3

R3

+

+

+

+

+

K1

K1

K2

K2

K3

= 4

= 16

= 24

= 16

= 24

To solve the five equations, then we proceed as follows. If R1 = 0, then

R1+ K1= 4

0 + K1 = 4

K1 = 4

R2 + K1 = 16

R2 + 4 = 16

R2 = 12

R2 + K2 = 24

12 + K2 = 24

K2 = 12

R3 + K2 = 16

R3 + 12 = 16

R3 = 4

R3 + K3 = 24

4 + K3 = 24

K3 = 20

Project A

Project B

Project C

Plant Capacit

y

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102 41

56

82

77

R2 =12

R3 = 4

Ri

Kj K1 = 4 K2 = 12 K3 = 20

R1 = 0

215

215

56

16

66

36

41

4 8 8

16

24

16

16

24

8

WA WB WC

XA XB XC

YA YB YC

From

To

Unused Square

C ij – Ri – Kj Improvement Index

12 C12 - R1 – K2

8 – 0 – 12- 4

13 C13 - R1 - K3

8 – 0 - 20-12

23 C23 – R2 - K3

16 – 12 - 20-16

31 C31 – R3 – K1

8 – 4 - 40

Traced a close path for the cell having the largest negative improvement index

Placed plus and minus signs at alternative corners of the path beginning with a plus sign at the unused square.

The smallest stone in a negative position on the close path indicates the quantity that can be assigned to the unused square.

Finally, the improvement indices for the new solution are calculated.

Project A

Project B

Project C

Plant Capacit

y

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102 41

56

82

77

R2

R3

Ri

Kj K1 K2 K3

R1

215

215

56

16

25

77

41

4 8 8

16

24

16

16

24

8

WA WB WC

XA XB XC

YA YB YC

From

To

Stone square 11:

R1+ K1= 4

0 + K1 = 4

K1 = 4

Stone square 21:

R2+ K1= 16

R2+ 4 = 16

R2 = 12

Stone square 22:

R2 + K2 = 24

12 + K2 = 24

K2 = 12

Stone square 23:

R2 + K3 = 16

12 + K3 = 16

K3 = 4

Stone Square 32:

R3 + K2 = 16

R3 + 12 = 16

R3 = 4

Unused Square


12 C12 - R1 – K2

8 – 0 - 12- 4

13 C13 - R1 – K3

8 – 0 - 4+ 4

31 C31 – R3 – K1

8 – 4 - 40

33 C33 – R3 – K3

24 – 4 - 4+16

Project A

Project B

Project C

Plant Capacit

y

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102 41

56

82

77

R2 =12

R3 = 8

Ri

Kj K1 = 4 K2 = 8 K3 = 4

R1 = 0

215

215

31

41

25

77

41

4 8 8

16

24

16

16

24

8

WA WB WC

XA XB XC

YA YB YC

From

To

Unused Square


13 C13 - R1 – K3

8 – 0 - 4+ 4

22 C22 – R2 – K2

24 – 12 - 8+ 4

31 C31 – R3 – K1

8 – 8 - 4- 4

33 C33 – R3 – K3

24 – 8 - 4+12

Project A

Project B

Project C

Plant Capacit

y

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102 41

56

82

77

R2

R3

Ri

Kj K1 K2 K3

R1

215

215

31

41

56

46

41

4 8 8

16

24

16

16

24

8

WA WB WC

XA XB XC

YA YB YC

From

To

Stone square 12:

R1+ K2= 8

0 + K2 = 8

K2 = 8

Stone square 21:

R2+ K1= 16

R2 + 0 = 16

R2 = 16

Stone square 23:

R2 + K3 = 16

16 + K3 = 16

K3 = 0

Stone square 31:

R3 + K1 = 8

R3 + 0 = 8

R3 = 8

Stone Square 32:

R3 + K2 = 16

8 + K2 = 16

K2 = 8

Unused Square


11 C11 - R1 – K1

4 – 0 - 0+ 4

13 C13 – R1 – K3

8 – 0 - 0+ 8

22 C22 – R2 – K2

24 – 16 - 80

33 C33 – R3 – K3

24 – 8 - 0+16

Is this the Optimal Solution?

1. For each solution, compute the R and K values for the table

2. Calculate the improvement indices for all unused squares

3. Select the unused square with the most negative index

4. Trace the closed path for the unused square having the most negative index.

5. Develop an improved solution

6. Repeat steps 1 to 5 until an optimal solution has been found.

Demand Less than Supply

Considering the original Bulacan Gravel Company problem, suppose that plant W has a capacity of 76 truckloads for week rather than 56. The company would be able to supply 235 truckloads per week. However, the project requirements remain the same .

Project A

Project B

Project C

Dummy D

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102 41 20

Plant Capacit

y

76

82

77

72

82

16

41

4 8 8

16

24

16

16

24

8

WA WB WC

XA XB XC

YA YB YC

From

To

0

4

20

0

0

235

235

Total Cost:

72 x P4 = P 288

4 x P8 = 32

82 x P24= 1,968

16 x P16 = 256

41 x P24 = 984

20x P0 = 0

P 3,528

Project A

Project B

Project C

Dummy D

Project Requirem

ents

Plant Y

Plant X

Plant W

72 102 41 20

Plant Capacit

y

76

82

77

76

21

72

5

4 8 8

16

24

16

16

24

8

WA WB WC

XA XB XC

YA YB YC

From

To

0

20

0

0

235

235

Total Cost:

76 x P8 = P 608

21 x P24= 504

41 x P16= 656

20 x P0 = 0

72 x P8 = 576

5x P16 = 80

P 2,424

41

Demand Greater than Supply Assume that project A will require 10 additional truckloads per week

and that project C estimates additional requirements of 20 truckloads. The total project requirements now would be equal to 245 truckloads, as opposed to the 215 available from the plants.

Total Cost:

56 x P4 = P224

26 x P16= 416

56 x P24= 1,344

46 x P16= 736

31 x P24= 744

30 x P 0= 0

P 3,464

Total Cost:

56 x P 8 = P 448

21 x P16 = 336

61 x P16 = 976

61 x P 8 = 488

16 x P 16 = 256

30 x P 0 = 0

P 2,504

There maybe an excessive number of stone squares in a solution; the number of stone squares is greater than the number of rim requirements minus 1.

There may be an insufficient number of stone squares in a solution.

DEGENERACY IN ESTABLISHING AN INITIAL SOLUTION

From


Plant Capacity

Project Requirem

ents

Plant Y

Plant X

Plant W

35 45 35

55

25

35

115115

4 8

16

824

16

8

24

16

WA WB WC

XA

YA

XB XC

YB YC

35

16 25

35

20

From


Plant Capacity

Project Requirem

ents

Plant Y

Plant X

Plant W

35 45 35

55

25

35

115115

4 8

16

824

16

8

24

16

WA WB WC

XA

YA

XB XC

YB YC

35

16 25

35

20

0

From

ToA B C

Plant Capacity

Project Requirem

ents

Y

X

W

30 25 45

25

35

40

100100

9 8

6

7 9

4

5

8

6

TRANSPORTATION TABLE FOR SUPER FERRY SHIPPING

Shipping Assignments

Quantity Shipped

x Unit Cost

= Total Cost

WC 25 5 P125

WX 15 6 90

WC 20 4 80

YAYB

1525

76

105150

P550Total Transportation Cost

0


Quantity Shipped

x Unit Cost

= Total Cost

WA 25 9 P225

XA 5 6 30

XBXC

255

84

20020

YC 40 9 360P835Total Transportation Cost

0


Quantity Shipped

X Unit Cost

= Total Cost

WA 15 9 P135

WC 10 5 50

XC 35 4 140

YAYB

1525

76

105150

P580Total Transportation Cost

0

Opportunity Costs for the first allocation

First Allocation using the VAM (Vogel Approximation Method)

Row W

Row X

0

Opportunity costs for the

Second allocation

Second Allocation

Using the VAM

0

Opportunity costs for the

Third allocation

Second Allocation

Using the VAM

•THE ASSIGNMENT PROBLEM

The Meycauayan Machine Shop does custom metalworking for a number of local plants. Meycauayan currently has three jobs to be done (let us symbolize them A, B and C). Meycauayan also has three machines on which to do the work (X, Y, and Z).

Anyone of the jobs can be processed completely on any of the machines. Furthermore, the cost of processing any job on any machine is known. The assignment of jobs to machine must be on one-to-one basis; that is, each job must be assigned exclusively to one and only one machine. T jobs to the objective is to assign the jobs to the machines so as to minimize the cost.

Job X Y ZTotal

A P25 P31 P35 91

B 15 20 24 59

C 22 19 17 58

Total 62 70 76 208

Machine

•THE ASSIGNMENT PROBLEM

STEP 1: Determine the opportunity- cost table

Job X Y Z

A 10 12 18

B 0 1 7

C 7 0 0

Column X Column Y Column Z

25-15=10 31-19 = 12 35 – 17 = 18

15-15 =0 20 – 19 = 1 24 – 17 = 7

22-15 = 7 19 – 19 = 0 17 – 17 = 0

Computations

Step 1; part a

0

X Y Z

Row A 10 – 10 = 0 12 – 10 = 2 18 – 10 = 8

Row B 0 – 0 = 0 1 – 0 = 1 7 – 0 = 7

Row C 7 – 0 = 7 0 – 0 = 0 0 – 0 = 0

STEP 2: Determine whether an optimal assignment can be made

Computations:Step 1; part b

Line 2

Line 1

0

STEP 3: Revise the total opportunity-cost table

0 2 – 1 = 1 8 – 1 = 7

0 1 – 1 = 0 7 – 1 = 6

8 0 0

a) Subtract

lowest number

from all uncovered

numberLine 1

Line 2

7 + 1 =

b) Add same smallest number to

numbers lying at the intersection

of two lines

Computations:

Job X Y Z

A 0 1 7

B 0 0 6

C 8 0 0

revised opportunity

cost table

Line 2

Line 1

Line 3

0

Total Cost:

Assignment Cost

A to X P25

B to Y 20

C to Z 17

P62

•MAXIMIZATION PROBLEMS

0

TRANSPORTATION FOR MAXIMIZATION PROBLEMS Juan dela Cruz is Marketing Vice President with the Malolos Company. Malolos is

planning to expand its sales of computer software into new towns – Angat, Norzagaray,and Baliuag. It has been determined that 20, 15 , and 30 salespersons will be requiredto service each of the three areas, respectively. The company recently hired 65 newsalespersons to cover these areas, and based on their experience and prior salesperformance, the new hires have been classified as Type A, B, or C salesperson.Depending upon the regions to which each of the three types are assigned, Malolosestimates the following annual revenues per salesperson:

Dave wishes to determine how many salespersons of each type to assign to the threeregions so that total annual revenues will be maximized.

Type Number of Salespersons

availableAngat Norzagaray Baliuag

A 24 P100,000 120,000 130,000

B 27 90,000 106,000 126,000

C 14 84,000 98,000 120,000

Annual Revenues

0

evaluation of unused squares

0

Salesperson Type

Territory

Number

Assigned

Annual Revenue

A Angat 9 P900, 000

A Norzagaray

15 1,800,000

B Angat 11 990,000

B Baliuag 16 2,016,000

C Baliuag 14 1,680,000P7,386,000

0

STEP 1: Select the highest and second-highest revenuealternatives from among those not already allocated. Thedifference between these two revenues will be theopportunity cost for the row or column.

STEP 2: Scan these opportunity-cost figures and identify therow or columns with the largest opportunity cost.

STEP 3: Allocate as many units as possible to this row orcolumn in the square with the greatest revenue.

0

ASSIGNMENT FOR METHOD OF

MAXIMIZATION PROBLEMS

Josefa Santos manages the Aves Car Rental Agency. This year, sheplans to purchase five new automobiles to replace five oldervehicles. The older vehicles are to be sold at auction. Josefa hassolicited bid from five individuals, each of whom wishes to purchaseonly one vehicle but has agreed to make a sealed bid on each ofthe five. The bids are as follows:

Buyer Ford Honda Kia Mitsubishi Toyota

Amalia P 3,000 P 2,500 P 3,300 P 2,600 P 3,100

Berto 3,500 3,000 2,800 2,800 3,300

Carlos 2,800 2,900 3,900 2,300 3,600

Dolores 3,300 3,100 3,400 2,900 3,500

Edgardo 2,800 3,500 3,600 2,900 3,000

Automobile

0


Amalia 500 1000 600 300 500

Berto 0 500 1100 100 300

Carlos 700 600 0 600 0

Dolores 200 400 500 0 100

Edgardo 700 0 300 0 600


Amalia 200 700 200 0 100

Berto 0 500 1000 100 200

Carlos 800 700 0 700 0

Dolores 200 400 400 0 0

Edgardo 700 0 200 0 500

Automobile

Automobile

•REGRET VALUES FOR AUTOMOBILE

•OPTIMAL ASSIGNMENT OF BID AWARDS

0

Buyer Bid Accepted Bid Price

Amalia Mitsubishi P 2,600

Berto Ford 3,500

Carlos Kia 3,900

Dolores Toyota 3,500

Edgardo Honda 3,500P 17,000

Reference: Quantitative Approaches to Management, 8th edition; Richard I. Levin, et. al

•OPTIMAL SOLUTION TO Aves CAR RENTAL AGENCY

Thank You

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